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Hidden Markov Models. 1. 2. K. …. Outline. Hidden Markov Models – Formalism The Three Basic Problems of HMMs Solutions Applications of HMMs for Automatic Speech Recognition (ASR). Example: The Dishonest Casino. A casino has two dice: Fair die P(1) = P(2) = P(3) = P(5) = P(6) = 1/6

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## Hidden Markov Models

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**Hidden Markov Models**1 2 K …**Outline**• Hidden Markov Models – Formalism • The Three Basic Problems of HMMs • Solutions • Applications of HMMs for Automatic Speech Recognition (ASR)**Example: The Dishonest Casino**A casino has two dice: • Fair die P(1) = P(2) = P(3) = P(5) = P(6) = 1/6 • Loaded die P(1) = P(2) = P(3) = P(4) = P(5) = 1/10 P(6) = 1/2 Casino player switches back-&-forth between fair and loaded die once in a while Game: • You bet $1 • You roll (always with a fair die) • Casino player rolls (maybe with fair die, maybe with loaded die) • Highest number wins $2**Question # 1 – Evaluation**GIVEN A sequence of rolls by the casino player 12455264621461461361366616646616366163661636165 QUESTION How likely is this sequence, given our model of how the casino works? This is the EVALUATION problem in HMMs**Question # 2 – Decoding**GIVEN A sequence of rolls by the casino player 12455264621461461361366616646616366163661636165 QUESTION What portion of the sequence was generated with the fair die, and what portion with the loaded die? This is the DECODING question in HMMs**Question # 3 – Learning**GIVEN A sequence of rolls by the casino player 12455264621461461361366616646616366163661636165 QUESTION How “loaded” is the loaded die? How “fair” is the fair die? How often does the casino player change from fair to loaded, and back? This is the LEARNING question in HMMs**The dishonest casino model**0.05 0.95 0.95 FAIR LOADED P(1|F) = 1/6 P(2|F) = 1/6 P(3|F) = 1/6 P(4|F) = 1/6 P(5|F) = 1/6 P(6|F) = 1/6 P(1|L) = 1/10 P(2|L) = 1/10 P(3|L) = 1/10 P(4|L) = 1/10 P(5|L) = 1/10 P(6|L) = 1/2 0.05**Example: the dishonest casino**Let the sequence of rolls be: O = 1, 2, 1, 5, 6, 2, 1, 6, 2, 4 Then, what is the likelihood of X= Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair, Fair? (say initial probs P(t=0,Fair) = ½, P(t=0,Loaded)= ½) ½ P(1 | Fair) P(Fair | Fair) P(2 | Fair) P(Fair | Fair) … P(4 | Fair) = ½ (1/6)10 (0.95)9 = .00000000521158647211 = 0.5 10-9**Example: the dishonest casino**So, the likelihood the die is fair in all this run is just 0.521 10-9 OK, but what is the likelihood of X= Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded, Loaded? ½ P(1 | Loaded) P(Loaded, Loaded) … P(4 | Loaded) = ½ (1/10)8 (1/2)2 (0.95)9 = .00000000078781176215 = 7.9 10-10 Therefore, it is after all 6.59 times more likely that the die is fair all the way, than that it is loaded all the way.**Example: the dishonest casino**Let the sequence of rolls be: O = 1, 6, 6, 5, 6, 2, 6, 6, 3, 6 Now, what is the likelihood X = F, F, …, F? ½ (1/6)10 (0.95)9 = 0.5 10-9, same as before What is the likelihood X= L, L, …, L? ½ (1/10)4 (1/2)6 (0.95)9 = .00000049238235134735 = 0.5 10-7 So, it is 100 times more likely the die is loaded**HMM Timeline**Arrows indicate probabilistic dependencies. x’s are hidden states, each dependent only on the previous state. The Markov assumption holds for the state sequence. o’s are observations, dependent only on their corresponding hidden state. x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT time**HMM Formalism**An HMM can be specified by 3 matrices {P, A, B}: P = {pi} are the initial state probabilities A = {aij} are the state transition probabilities = Pr(xj|xi) B = {bik} are the observation probabilities = Pr(ok|xi) x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Generating a sequence by the model**1 1 1 1 … 2 2 2 2 … … … … … N N K N … Given a HMM, we can generate a sequence of length n as follows: • Start at state xi according to prob i • Emit letter o1 according to prob bi(o1) • Go to state xj according to prob aij • … until emitting oT 1 2 2 2 0 N b2o1 o1 o2 o3 oT**The three main questions on HMMs**• Evaluation GIVEN a HMM , and a sequence O, FIND Prob[ O | ] • Decoding GIVEN a HMM , and a sequence O, FIND the sequence X of states that maximizes P[X | O, ] • Learning GIVEN a sequence O, FIND a model with parameters , A and B that maximize P[ O | ]**Problem 1: Evaluation**Find the likelihood a sequence is generated by the model**Probability of an Observation**oT o1 ot-1 ot ot+1 Given an observation sequence and a model, compute the probability of the observation sequence**Probability of an Observation**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT Let X = x1 … xt be the state sequence.**HMM – Evaluation (cont.)**• Why isn’t it efficient? • For a given state sequence of length T we have about 2T calculations • Let N be the number of states in the graph. • There are NT possible state sequences. • Complexity : O(2TNT ) • Can be done more efficiently by the forward-backward (F-B) procedure.**The Forward Procedure (Prefix Probs)**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT The probability of being in state i after generating the first t observations.**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**Forward Procedure**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT**The Forward Procedure**Initialization: Iteration: Termination: Computational Complexity: O(N2T)**Another Version: The Backward Procedure (Suffix Probs)**x1 xt-1 xt xt+1 xT o1 ot-1 ot ot+1 oT Probability of the rest of the states given the first state**Problem 2: Decoding**Find the best state sequence**Decoding**• Given an HMM and a new sequence of observations, find the most probable sequence of hidden states that generated these observations: • In general, there is an exponential number of possible sequences. • Use dynamic programming to reduce search space to O(n2T).**Viterbi Algorithm**x1 xt-1 j o1 ot-1 ot ot+1 oT The state sequence which maximizes the probability of seeing the observations up to time t-1, landing in state j, and seeing the observation at time t.**Viterbi Algorithm**x1 xt-1 j o1 ot-1 ot ot+1 oT Initialization**Viterbi Algorithm**o1 ot-1 ot ot+1 oT x1 xt-1 xt xt+1 Recursion Prob. of ML state Name of ML state**Viterbi Algorithm**o1 ot-1 ot ot+1 oT x1 xt-1 xt xt+1 xT Termination “Read out” the most likely state sequence, working backwards.**Problem 3: Learning**Re-estimate the parameters of the model based on training data**Learning by Parameter Estimation:**• Goal : Given an observation sequence, find the model that is most likely to produce that sequence. • Problem: We don’t know the relative frequencies of hidden visited states. • No analytical solution is known for HMMs. • We will approach the solution by successive approximations.**The Baum-Welch Algorithm**• Find the expected frequencies of possible values of the hidden variables. • Compute the maximum likelihood distributions of the hidden variables (by normalizing, as usual for MLE). • Repeat until “convergence.” • This is the Expectation-Maximization (EM) algorithm for parameter estimation. • Applicable to any stochastic process, in theory. • Special case for HMMs is called the Baum-Welch algorithm.**Arc and State Probabilities**o1 ot-1 ot ot+1 oT A A A A B B B B B Probability of traversing an arc From state i (at time t) to state j (at time t+1) Probability of being in state i at time t.**Aggregation and Normalization**o1 ot-1 ot ot+1 oT A A A A B B B B B Now we can compute the new MLEs of the model parameters.**The Baum-Welch Algorithm**• Initialize A,B and (Pick the best-guess for model parameters or arbitrary) • Repeat • Calculate and • Calculate and • Estimate , and • Until the changes are small enough**The Baum-Welch Algorithm – Comments**Time Complexity: # iterations O(N2T) • Guaranteed to increase the (log) likelihood of the model P( | O) = P(O, ) / P(O) = P(O | ) / ( P(O) P() ) • Not guaranteed to find globally best parameters Converges to local optimum, depending on initial conditions • Too many parameters / too large model - Overtraining

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